Since Hendrix and Larson was not at 5M, let's say 7 mill then (between 6 and 8 million).

I looked up Dr Seuss on wikipedia*, and found that he wrote the books about Horton. I think these books was the material for the Horton films (is this right?), quite popular films. He therefore should be at about 8 million (between 7 and 9 million)?

*Is this cheating? I did not look up the list itself, but I had to look up some facts about the guy to make a guess. It this is considered cheating, please ignore my guess

Since Hendrix and Larson was not at 5M, let's say 7 mill then (between 6 and 8 million).

I looked up Dr Seuss on wikipedia*, and found that he wrote the books about Horton. I think these books was the material for the Horton films (is this right?), quite popular films. He therefore should be at about 8 million (between 7 and 9 million)?

*Is this cheating? I did not look up the list itself, but I had to look up some facts about the guy to make a guess. It this is considered cheating, please ignore my guess

Correct on all accounts! And you had an unbeatable lead even without Dr Seuss, so I'm happy to give you a pass on that.

Imagine yourself as the detective in a forgery case. Laying in a row before you are 11 coins. You know that three of them are false and 8 are genuine. You also know that the three forged coins are laying together, side by side, somewhere in the row. The false coins are heavier than the genuine one. How many coins (at a minimum) do you have to weigh before you can point out the three false coins.

Imagine yourself as the detective in a forgery case. Laying in a row before you are 11 coins. You know that three of them are false and 8 are genuine. You also know that the three forged coins are laying together, side by side, somewhere in the row. The false coins are heavier than the genuine one. How many coins (at a minimum) do you have to weigh before you can point out the three false coins.

number the coins 1..11.
Weigh coin 3 against coin 9.
If they are the same weight, we know that they are both good coins (because they are more than three coins apart, they can't both be bad coins). So coins 1,2,10,11 must also be good coins (because the three bad coins are together). This leaves us with the five coins 4 ..8 to examine.
If they are not the same weight, we know which one is the bad coin, because it is heavier. So we now know that the bad coins are in the five coins 1..5 or 7..11.

So after one weighing of two coins we've identified five coins that must contain the three bad coins.

Weigh the two coins at the ends of the five coins (i.e. 1 and 5 or 4 and 8 or 7 and 11).

If the coins are the same weight, the bad coins are in the middle of the five. If one of them is bad, the bad coins are at that end of the row of five.

You are quite right. When I made the question, I had in mind a weight that measured the wieght of just one coin, you imagined to weigh them against each other, but that doesn't matter. The Grand Mouse has the floor

I thought of a better solution, if you know beforehand the weight of a good coin.

Weigh the middle one.
If it is a bad coin, the three bad coins must be in the five coins in the middle. We can weigh just two more coins and find all the bad ones.
If it is a good coin, the bad coins must be in the first five or the last five. And we know how to do that with weighing just two coins.

Bad coins found after weighing just three coins!

But this solution does require that we know how much a good coin weighs.